`{:("Column-I","Column-II"),(A.f(x) = (1)/(sqrt(x -2)),p.lim_(x to 0)f(x) =1),(B. f(x) = (3x - "sin"x)/(x + "sin" x), q. lim_(x to 0)f(x) = 0),(C.f(x) = x "sin"(pi)/(x) f(0)=0,r.lim_(x to oo) f(x) = 0),(f(x) = tan^(-1) (1)/(x),s.lim_(x to 0) "does not exist"):}`

Answer

To solve the problem, we need to match the functions in Column I with their corresponding limits in Column II. Let's analyze each function step by step.

### Step 1: Analyze Function A
**Function:** \( f(x) = \frac{1}{\sqrt{x - 2}} \)

**Limit as \( x \to 0 \):**
- Substitute \( x = 0 \):
\[ f(0) = \frac{1}{\sqrt{0 - 2}} = \frac{1}{\sqrt{-2}} \text{ (imaginary number)} \]
- Therefore, the limit does not exist.

**Limit as \( x \to \infty \):**
- Substitute \( x \to \infty \):
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{1}{\sqrt{x - 2}} = \frac{1}{\sqrt{\infty}} = 0 \]

**Conclusion for A:**
- \( \lim_{x \to 0} f(x) \) does not exist.
- \( \lim_{x \to \infty} f(x) = 0 \) matches with \( r \).

### Step 2: Analyze Function B
**Function:** \( f(x) = \frac{3x - \sin x}{x + \sin x} \)

**Limit as \( x \to 0 \):**
- Substitute \( x = 0 \):
\[ f(0) = \frac{3(0) - \sin(0)}{0 + \sin(0)} = \frac{0}{0} \text{ (indeterminate form)} \]
- Apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{3 - \cos x}{1 + \cos x} \]
- Substitute \( x = 0 \):
\[ = \frac{3 - 1}{1 + 1} = \frac{2}{2} = 1 \]

**Conclusion for B:**
- \( \lim_{x \to 0} f(x) = 1 \) matches with \( p \).

### Step 3: Analyze Function C
**Function:** \( f(x) = x \sin\left(\frac{\pi}{x}\right) \), with \( f(0) = 0 \)

**Limit as \( x \to 0 \):**
- Substitute \( x \to 0 \):
\[ \lim_{x \to 0} x \sin\left(\frac{\pi}{x}\right) \]
- Since \( \sin\left(\frac{\pi}{x}\right) \) oscillates between -1 and 1, we have:
\[ -x \leq x \sin\left(\frac{\pi}{x}\right) \leq x \]
- As \( x \to 0 \), both bounds approach 0.
- By the Squeeze Theorem:
\[ \lim_{x \to 0} x \sin\left(\frac{\pi}{x}\right) = 0 \]

**Conclusion for C:**
- \( \lim_{x \to 0} f(x) = 0 \) matches with \( q \).

### Step 4: Analyze Function D
**Function:** \( f(x) = \tan^{-1}\left(\frac{1}{x}\right) \)

**Limit as \( x \to 0 \):**
- As \( x \to 0^+ \):
\[ \tan^{-1}\left(\frac{1}{0^+}\right) = \tan^{-1}(\infty) = \frac{\pi}{2} \]
- As \( x \to 0^- \):
\[ \tan^{-1}\left(\frac{1}{0^-}\right) = \tan^{-1}(-\infty) = -\frac{\pi}{2} \]
- Since the left-hand limit and right-hand limit do not match, the limit does not exist.

**Conclusion for D:**
- \( \lim_{x \to 0} f(x) \) does not exist, matches with \( s \).

### Final Matching:
- A matches with \( r \): \( \lim_{x \to \infty} f(x) = 0 \)
- B matches with \( p \): \( \lim_{x \to 0} f(x) = 1 \)
- C matches with \( q \): \( \lim_{x \to 0} f(x) = 0 \)
- D matches with \( s \): \( \lim_{x \to 0} f(x) \) does not exist.

### Summary of Results:
- A → r
- B → p
- C → q
- D → s